Analytic properties of Euler sums Introduction
As far as I know this topic has not been discussed before. I have found interesting results which I wish to share with you in the standard MSE manner by asking questions.
Consider the Euler sum
$$S(1,s) =\sum _{k=1}^{\infty } \frac{H_k}{k^s}\tag{1}$$
where $H_k = \sum_{j=1}^k 1/j$ is the harmonic number and $s$ is a parameter which in the field of Euler sums is considered to assume integer values $\ge 2$.
Here we let $s$ be a complex number and ask for the analytic properties of $S(1,s)$ as a function of $s$.
In parallel and for comparison we consider the similar sum with the harmonic number replaced by unitiy. This is the well known zeta function studied first by Euler
$$\zeta(s) = \sum _{k=1}^{\infty } \frac{1}{k^s}\tag{2}$$
The analytic properties of this function have been studied first by Bernhard Riemann in his famous paper on the distribution of prime numbers of 1859 [1].
The zeta function is analytic in the whole complex $s$ plane except for a simple pole at $s=1$ with residue $1$. It has trivial simple zeroes at negative even integers and non trivial zeroes on the line $\Re[s] = 1/2$ possibly even all non trivial zeroes are located on this line (Riemann's hypothesis) [2],[3].
Question 1
What are the singularities of $S(1,s)$ in the complex s-plane? What can be said about the zeroes? Compare the results with those of $\zeta(s)$
First solution steps: derive an integral representation of $S(1,s)$, and split is into an holomorphic and a meromorphic part. Study the meromorphic part to find the singularities.
Question 2
$S(1,q)$ has a well known closed form for integer argument $q = 2, 3, 4, ..$ which was already found by Euler. 
It is given by
$$S_E(1,q) = \left(\frac{q}{2}+1\right) \zeta (q+1)-\frac{1}{2} \sum _{k=1}^{q-2} \zeta (k+1) \zeta (q-k)\tag{3}$$
The question asks for a possible closed form for real $q \ge 2$
First solution steps: Here I have no idea how to find a better solution than the integral representation.
Question 3
Consider the analogue questions 1 and 2 also for the general linear harmonic sum including the generating function via the parameter $x$
$$S_H(p, q, x)=\sum _{k=1}^{\infty } \frac{x^k H_k^{(p)}}{k^q}\tag{4}$$
Here $H_k^{(p)}=\sum_{n=1}^k 1/n^p$ is the generalized harmonic number.
As there are three quantities in the game, to be definite, we consider some simplifying cases.
Case 1: $x=1$ and $x=-1$ (alternating sum) for fixed $p$ as a function of $q$
Case 2: fixed $x$ and $p=q$ as a function of $p$. Notice that there is an analogous formula to (3) for integer $p$
Case 3: $X$ fixed somewhere within the interval $(-1,1)$, e.g. $x=\frac{1}{2}$ and, say, $p=1$.
References
[1] http://www.claymath.org/sites/default/files/zeta.pdf (Bernhard Riemann, Über die Anzahl der Primzahlen unter einer gegebenen Größe, Berlin November 1859)
[2] https://en.wikipedia.org/wiki/Riemann_zeta_function
[3] https://de.wikipedia.org/wiki/Riemannsche_%CE%B6-Funktion 
 A: Let us start collecting something.
$$\begin{eqnarray*} S(1,s) = \sum_{n\geq 1}\frac{\psi(n+1)+\gamma}{n^s} &=& \sum_{n\geq 1}\frac{1}{n^s}\left[\log n+\gamma+\frac{1}{2n}-\sum_{j\geq 1}\frac{B_{2j}}{2j n^{2j}}\right]\\&=&\sum_{n\geq 1}\frac{1}{n^s}\left[\log n+\gamma+\frac{1}{2n}-\sum_{j\geq 1}\frac{(-1)^{j+1}(2j)! \zeta(2j)}{j (2\pi n)^{2j}}\right]\\&\stackrel{\mathcal{L}^{-1}}{=}&\int_{0}^{+\infty}\frac{1}{e^u-1}\left(\frac{u^{s-1}}{\Gamma(s)}(\psi(s)-\log u+\gamma)-\sum_{j\geq 1}\frac{(-1)^{j+1}(2j)!\zeta(2j)u^{s+2j-1}}{j(2\pi)^{2j}\Gamma(2j+s)}\right)\,du\end{eqnarray*}$$
and if $s\in\{2,3,4,\ldots\}$ the series on $j$ simplifies in terms of the $\log\Gamma$ function, such that an explicit evaluation of the last integral is allowed by the residue theorem and the reflection formula. The similarity between $S(1,s)$ and $-\zeta'(s)$ is encoded by the term $\frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{u^{s-1}\log(u)}{e^u-1}$, and the integral representation allows an analytic continuation of $S(1,s)$ to the complex plane. On the other hand, due to the presence of many spurious terms it is not clear how the roots of $S(1,s)$ and $-\zeta'(s)$ are related.
Still $S(1,s)$ has a reflection formula. We may derive it in the same way we derive the reflection formula for the $\zeta$ function, i.e. by enforcing the substitution $u\mapsto v^2$ in the integral representation and by exploiting the Poisson summation formula. The involved computations are pretty heavy so I need some time for performing an update of this (partial) answer.
